The Properties of the Tangent Function

Determine the properties of the function |f(x)=-\tan\left(\frac{1}{2}(x-1)\right)+\sqrt{2}| .

It may be useful to plot a graph of the function.

• The coordinates of the inflection point are |(h,k)=(1,\sqrt{2})|.

• The period of the function is:||\displaystyle P = \frac{\pi}{\mid b \mid} = \frac{\pi}{\frac{1}{2}} = 2\pi||

• The equation of the asymptotes are:||\begin{align} x &= \left(h + \frac{P}{2}\right) + n P\\ &= \left(1+\frac{2\pi}{2}\right) +n (2\pi)\\ &= (1+\pi) + 2\pi n \end{align}|| where |n \in \mathbb{Z}| and |P| is the period.

• The domain of the function is: |\mathbb{R} \backslash \lbrace (1+\pi) + 2 \pi n \rbrace| where |n \in \mathbb{Z}| and |P| is the period.

• The range of the function is the set of real numbers, i.e. |\mathbb{R}.|

• The interval: from the values of |a| and |b,| the function must be decreasing, since the product |a b| is negative |\left(-1 \times \frac{1}{2} <0\right).| The graph confirms it. 

• The zeroes of the function are calculated by replacing |f(x)| by |0.| ||\begin{align}0 &= -\tan\left(\frac{1}{2}(x-1)\right)+\sqrt{2}\\-\sqrt{2} &= - \tan\left(\frac{1}{2}(x-1)\right)\\ \sqrt{2} &= \tan\left(\frac{1}{2}(x-1)\right)\end{align}|| At this step, check what angle the tangent is |\sqrt{2}.| Look at the interval angle |[0,\pi].| The value is |0{.}955.|
  
  Thus, the interior of the tangent function is equal to |0{.}955.| ||\begin{align}0{.}955 &= \frac{1}{2}(x-1)\\1{.}91 &= x-1\\2{.}91 &= x\end{align}||The zero of the function in the cycle is |2{.}91.|
  
  The general expression for the function’s zeroes is |x=2{.}91 + 2\pi n| where |n \in \mathbb{Z}.|

• The positive and negative intervals: the function is positive on the interval |(1-\pi + 2\pi n,\ 2{.}91 + 2 \pi n]| and negative on the interval |[2{.}91 + 2 \pi n,\ 1+ \pi + 2 \pi n)| where |n \in \mathbb{Z}|.
  Be careful not to include asymptotes.

• The function has no extrema.